Let's first prove the identity cos θ= cot θcsc θ. We will use basic identities to show how the right-hand side is equal to the left-hand side.
\cos \theta \stackrel{?}= \dfrac{\cot \theta}{\csc \theta}
\cos \theta \stackrel{?}= \cot \theta\left(\dfrac{1}{\csc \theta}\right)
cot(θ) = cos(θ)/sin(θ)
\cos \theta \stackrel{?}= {\color{#0000FF}{\dfrac{\cos \theta}{\sin \theta}}}\left(\dfrac{1}{\csc \theta}\right)
\cos \theta \stackrel{?}= \dfrac{\cos \theta}{\sin \theta}\left(\dfrac{1}{{\color{#0000FF}{\frac{1}{\sin \theta}}}}\right)
\cos \theta \stackrel{?}= \dfrac{\cos \theta}{\sin \theta}\left(\dfrac{\sin \theta}{1}\right)
\cos \theta \stackrel{?}= \dfrac{\cos \theta \cdot \sin \theta}{\sin \theta}
\cos \theta \stackrel{?}= \dfrac{\cos \theta \cdot \cancel{{\color{#FF0000}{\sin \theta}}}}{\cancel{{\color{#FF0000}{\sin \theta}}}}
cos θ = cos θ ✓
Having proved the identity cos θ = cot θcsc θ, we can prove the identity sec θ= csc θcot θ. This time, we will start on the left-hand side and arrive to the right-hand side.
\sec \theta \stackrel{?}=\dfrac{\csc \theta}{\cot \theta}
\dfrac{1}{\cos \theta} \stackrel{?}=\dfrac{\csc \theta}{\cot \theta}
\dfrac{1}{{\color{#0000FF}{\frac{\cot \theta}{\csc \theta}}}} \stackrel{?}=\dfrac{\csc \theta}{\cot \theta}
csc θ/cot θ = csc θ/cot θ ✓
By following similar procedures, we can prove that tan θ = 1cot θ.
\tan \theta \stackrel{?}= \dfrac{1}{\cot \theta}
\tan \theta \stackrel{?}= \dfrac{1}{{\color{#0000FF}{ \frac{\cos \theta}{\sin \theta} }}}
\tan \theta \stackrel{?}= \dfrac{\sin \theta}{\cos \theta}
tan θ = tan θ ✓
Let's now see how to rationalize the denominator of the 1- sqrt(10).
sin θ=1/- sqrt(10)
sin θ=- 1/sqrt(10)
sin θ=- sqrt(10)/sqrt(10)* sqrt(10)
sin θ=- sqrt(10)/(sqrt(10))^2
sin θ=- sqrt(10)/10
Let's now follow a similar procedure to rationalize the expression - 3- sqrt(10).
cos θ = - 3/- sqrt(10)
cos θ = 3/sqrt(10)
cos θ=3sqrt(10)/sqrt(10)* sqrt(10)
cos θ=3sqrt(10)/(sqrt(10))^2
cos θ=3sqrt(10)/10
